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</div>		<h1 class="firstHeading">Kinetic energy</h1>
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			<h3 id="siteSub">From Wikipedia, the free encyclopedia</h3>
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The cars of a <a href="http://en.wikipedia.org/wiki/Roller_coaster" title="Roller coaster">roller coaster</a>
reach their maximum kinetic energy when at the bottom of their path.
When they start rising, the kinetic energy begins to be converted to
gravitational potential energy, but the total amount of energy in the
system remains constant; assuming negligible <a href="http://en.wikipedia.org/wiki/Friction" title="Friction">friction</a> and other <a href="http://en.wikipedia.org/wiki/Energy_conversion" class="mw-redirect" title="Energy conversion">energy conversion</a> factors.</div>
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<p>The <b>kinetic energy</b> of an object is the extra <a href="http://en.wikipedia.org/wiki/Energy" title="Energy">energy</a> which it possesses due to its motion. It is defined as <i>the <a href="http://en.wikipedia.org/wiki/Mechanical_work" title="Mechanical work">work</a> needed to accelerate a body of a given mass from rest to its current velocity</i>. Having gained this energy during its <a href="http://en.wikipedia.org/wiki/Acceleration" title="Acceleration">acceleration</a>, the body maintains this kinetic energy unless its speed changes. <a href="http://en.wikipedia.org/wiki/Negative" title="Negative">Negative</a> work of the same magnitude would be required to return the body to a state of rest from that velocity.</p>
<table id="toc" class="toc" summary="Contents">
<tbody><tr>
<td>
<div id="toctitle">
<h2>Contents</h2>
 <span class="toctoggle">[<a href="javascript:toggleToc()" class="internal" id="togglelink">hide</a>]</span></div>
<ul>
<li class="toclevel-1"><a href="#Etymology"><span class="tocnumber">1</span> <span class="toctext">Etymology</span></a></li>
<li class="toclevel-1"><a href="#Introduction"><span class="tocnumber">2</span> <span class="toctext">Introduction</span></a>
<ul>
<li class="toclevel-2"><a href="#Calculations"><span class="tocnumber">2.1</span> <span class="toctext">Calculations</span></a></li>
</ul>
</li>
<li class="toclevel-1"><a href="#Newtonian_kinetic_energy"><span class="tocnumber">3</span> <span class="toctext">Newtonian kinetic energy</span></a>
<ul>
<li class="toclevel-2"><a href="#Kinetic_energy_of_rigid_bodies"><span class="tocnumber">3.1</span> <span class="toctext">Kinetic energy of rigid bodies</span></a></li>
<li class="toclevel-2"><a href="#Derivation_and_definition"><span class="tocnumber">3.2</span> <span class="toctext">Derivation and definition</span></a></li>
<li class="toclevel-2"><a href="#Kinetic_energy_of_systems"><span class="tocnumber">3.3</span> <span class="toctext">Kinetic energy of systems</span></a></li>
<li class="toclevel-2"><a href="#Rotating_bodies"><span class="tocnumber">3.4</span> <span class="toctext">Rotating bodies</span></a></li>
<li class="toclevel-2"><a href="#Rotation_in_systems"><span class="tocnumber">3.5</span> <span class="toctext">Rotation in systems</span></a></li>
</ul>
</li>
<li class="toclevel-1"><a href="#Relativistic_kinetic_energy_of_rigid_bodies"><span class="tocnumber">4</span> <span class="toctext">Relativistic kinetic energy of rigid bodies</span></a></li>
<li class="toclevel-1"><a href="#Quantum_mechanical_kinetic_energy_of_rigid_bodies"><span class="tocnumber">5</span> <span class="toctext">Quantum mechanical kinetic energy of rigid bodies</span></a></li>
<li class="toclevel-1"><a href="#Some_examples"><span class="tocnumber">6</span> <span class="toctext">Some examples</span></a></li>
<li class="toclevel-1"><a href="#See_also"><span class="tocnumber">7</span> <span class="toctext">See also</span></a></li>
<li class="toclevel-1"><a href="#Notes"><span class="tocnumber">8</span> <span class="toctext">Notes</span></a></li>
<li class="toclevel-1"><a href="#References"><span class="tocnumber">9</span> <span class="toctext">References</span></a></li>
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<p><a name="Etymology" id="Etymology"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=1" title="Edit section: Etymology">edit</a>]</span> <span class="mw-headline">Etymology</span></h2>
<p>The adjective "kinetic" to the noun <a href="http://en.wikipedia.org/wiki/Energy" title="Energy">energy</a> has its roots in the <a href="http://en.wikipedia.org/wiki/Ancient_Greek" title="Ancient Greek">Greek</a> word for "motion" (<a href="http://en.wikipedia.org/wiki/Kinesis" title="Kinesis">kinesis</a>). The terms <i>kinetic energy</i> and <i>work</i>
and their present scientific meanings date back to the mid 19th
century. Early understandings of these ideas can be attributed to <a href="http://en.wikipedia.org/wiki/Gaspard-Gustave_Coriolis" title="Gaspard-Gustave Coriolis">Gaspard-Gustave Coriolis</a> who in 1829 published the paper titled <i>Du Calcul de l'Effet des Machines</i> outlining the mathematics of kinetic energy.</p>
<p><a href="http://en.wikipedia.org/wiki/William_Thomson%2C_1st_Baron_Kelvin" title="William Thomson, 1st Baron Kelvin">William Thomson</a>, later Lord Kelvin, is given the credit for coining the term <i>kinetic energy</i> c. 1849.<sup class="noprint Template-Fact"><span title="This claim needs references to reliable sources&nbsp;since February 2008" style="white-space: nowrap;">[<i><a href="http://en.wikipedia.org/wiki/Wikipedia:Citation_needed" title="Wikipedia:Citation needed">citation needed</a></i>]</span></sup></p>
<p><a name="Introduction" id="Introduction"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=2" title="Edit section: Introduction">edit</a>]</span> <span class="mw-headline">Introduction</span></h2>
<dl>
<dd>
<div class="noprint relarticle mainarticle"><i>Main article: <a href="http://en.wikipedia.org/wiki/Energy" title="Energy">Energy</a></i></div>
</dd>
</dl>
<p>There are various forms of energy&nbsp;: <a href="http://en.wikipedia.org/wiki/Chemical_energy" class="mw-redirect" title="Chemical energy">chemical energy</a>, <a href="http://en.wikipedia.org/wiki/Heat" title="Heat">heat</a>, <a href="http://en.wikipedia.org/wiki/Electromagnetic_radiation" title="Electromagnetic radiation">electromagnetic radiation</a>, <a href="http://en.wikipedia.org/wiki/Potential_energy" title="Potential energy">potential energy</a> (gravitational, electric, elastic, etc.), <a href="http://en.wikipedia.org/wiki/Nuclear_energy" title="Nuclear energy">nuclear energy</a>, <a href="http://en.wikipedia.org/wiki/Rest_energy" title="Rest energy">rest energy</a>. These can be categorized in two main classes: <a href="http://en.wikipedia.org/wiki/Potential_energy" title="Potential energy">potential energy</a> and kinetic energy.</p>
<p>Kinetic energy can be best understood by examples that demonstrate
how it is transformed from other forms of energy and to the other
forms. For example, a cyclist will use <a href="http://en.wikipedia.org/wiki/Potential_energy" title="Potential energy">chemical energy</a>
that was provided by food to accelerate a bicycle to a chosen speed.
This speed can be maintained without further work, except to overcome
air-resistance and friction. The energy has been converted into the
energy of motion, known as <b>kinetic energy</b> but the process is not completely efficient and heat is also produced within the cyclist.</p>
<p>The kinetic energy in the moving <a href="http://en.wikipedia.org/wiki/Bicycle" title="Bicycle">bicycle</a> and the <a href="http://en.wikipedia.org/wiki/Cyclist" class="mw-redirect" title="Cyclist">cyclist</a>
can be converted to other forms. For example, the cyclist could
encounter a hill just high enough to coast up, so that the bicycle
comes to a complete halt at the top. The kinetic energy has now largely
been converted to gravitational potential energy that can be released
by freewheeling down the other side of the hill. (Since the bicycle
lost some of its energy to friction, it will never regain all of its
speed without further pedaling. Note that the energy is not destroyed;
it has only been converted to another form by friction.) Alternatively
the cyclist could connect a <a href="http://en.wikipedia.org/wiki/Electrical_generator" title="Electrical generator">dynamo</a>
to one of the wheels and also generate some electrical energy on the
descent. The bicycle would be traveling more slowly at the bottom of
the hill because some of the energy has been diverted into making
electrical power. Another possibility would be for the cyclist to apply
the brakes, in which case the kinetic energy would be dissipated
through friction as heat energy.</p>
<p>Like any physical quantity which is a function of velocity, the
kinetic energy of an object does not depend only on the inner nature of
that object. It also depends on the relationship between that object
and the observer (in physics an observer is formally defined by a
particular class of coordinate system called an <i><a href="http://en.wikipedia.org/wiki/Inertial_reference_frame" class="mw-redirect" title="Inertial reference frame">inertial reference frame</a></i>). Physical quantities like this are said to be not <i>invariant</i>. The kinetic energy is co-located with the object and contributes to its gravitational field.</p>
<p><a name="Calculations" id="Calculations"></a></p>
<h3><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=3" title="Edit section: Calculations">edit</a>]</span> <span class="mw-headline">Calculations</span></h3>
<p>There are several different equations that may be used to calculate
the kinetic energy of an object. In many cases they give almost the
same answer to well within measurable accuracy. Where they differ, the
choice of which to use is determined by the velocity of the body or its
size. Thus, if the object is moving at a velocity much smaller than the
speed of light, the <a href="http://en.wikipedia.org/wiki/Newtonian_mechanics" class="mw-redirect" title="Newtonian mechanics">Newtonian (classical) mechanics</a> will be sufficiently accurate; but if the speed is comparable to the speed of light, <a href="http://en.wikipedia.org/wiki/Special_relativity" title="Special relativity">relativity</a> starts to make significant differences to the result and should be used. If the size of the object is sub-atomic, the <a href="http://en.wikipedia.org/wiki/Quantum_mechanical" class="mw-redirect" title="Quantum mechanical">quantum mechanical</a> equation is most appropriate.</p>
<p><a name="Newtonian_kinetic_energy" id="Newtonian_kinetic_energy"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=4" title="Edit section: Newtonian kinetic energy">edit</a>]</span> <span class="mw-headline">Newtonian kinetic energy</span></h2>
<p><a name="Kinetic_energy_of_rigid_bodies" id="Kinetic_energy_of_rigid_bodies"></a></p>
<h3><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=5" title="Edit section: Kinetic energy of rigid bodies">edit</a>]</span> <span class="mw-headline">Kinetic energy of rigid bodies</span></h3>
<p>In <a href="http://en.wikipedia.org/wiki/Classical_mechanics" title="Classical mechanics">classical mechanics</a>, the kinetic energy of a "point object" (a body so small that its size can be ignored), or a non rotating <a href="http://en.wikipedia.org/wiki/Rigid_body" title="Rigid body">rigid body</a>, is given by the equation <img class="tex" alt="E_k = \begin{matrix} \frac{1}{2} \end{matrix} mv^2 " src="Kinetic_energy_files/231cfd9416f4736f5ee8d102ee84cb22.png"> where m is the mass and v is the speed of the body.</p>
<p>For example - one would calculate the kinetic energy of an 80 kg mass traveling at 18 meters per second (40 mph) as</p>
<dl>
<dd><img class="tex" alt="\begin{matrix} \frac{1}{2} \end{matrix} \cdot 80 \cdot 18^2 = 12,960 \ \mathrm{joules}" src="Kinetic_energy_files/e5fdfb17ce23a61f66295ff4e1ebe2b6.png">.</dd>
</dl>
<p>Note that the kinetic energy increases with the square of the speed.
This means, for example, that an object traveling twice as fast will
have four times as much kinetic energy. As a result of this, a car
traveling twice as fast requires four times as much distance to stop
(assuming a constant braking force. See <a href="http://en.wikipedia.org/wiki/Mechanical_work" title="Mechanical work">mechanical work</a>).</p>
<p>Thus, the kinetic energy can be calculated using the formula:</p>
<dl>
<dd><img class="tex" alt="E_k = \frac{1}{2}mv^2" src="Kinetic_energy_files/a38c32f3f00f593c1dc17692bc224c0f.png"></dd>
</dl>
<p>where:</p>
<dl>
<dd><i>E</i><sub>k</sub> is the kinetic energy in <a href="http://en.wikipedia.org/wiki/Joule" title="Joule">joules</a></dd>
<dd><i>m</i> is the mass in kilograms, and</dd>
<dd><i>v</i> is the speed in meters per second.</dd>
</dl>
<p>For the <i>translational kinetic energy</i> of a body with constant <a href="http://en.wikipedia.org/wiki/Mass" title="Mass">mass</a> <i>m</i>, whose <a href="http://en.wikipedia.org/wiki/Center_of_mass" title="Center of mass">center of mass</a> is moving in a straight line with speed <i>v</i>, as seen above is equal to</p>
<dl>
<dd><img class="tex" alt=" E_t = \begin{matrix} \frac{1}{2} \end{matrix} mv^2 " src="Kinetic_energy_files/099882d6f8b773d73d1039ef628e8a4e.png"></dd>
</dl>
<p>where:</p>
<dl>
<dd><i>m</i> is mass of the body</dd>
<dd><i>v</i> is speed of the <a href="http://en.wikipedia.org/wiki/Center_of_mass" title="Center of mass">center of mass</a> of the body.</dd>
</dl>
<p>Thus kinetic energy is a relative measure and no object can be said
to have a unique kinetic energy. A rocket engine could be seen to
transfer its energy to the rocket ship or to the exhaust stream
depending upon the chosen frame of reference. But the total energy of
the system, i.e. kinetic energy, fuel chemical energy, heat energy etc,
will be conserved regardless of the choice of measurement frame.</p>
<p>The kinetic energy of an object is related to its <a href="http://en.wikipedia.org/wiki/Momentum" title="Momentum">momentum</a> by the equation:</p>
<dl>
<dd><img class="tex" alt="E_k = \frac{p^2}{2m}" src="Kinetic_energy_files/1e8440b3262d1866bf61293634e00f19.png"></dd>
</dl>
<p><a name="Derivation_and_definition" id="Derivation_and_definition"></a></p>
<h3><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=6" title="Edit section: Derivation and definition">edit</a>]</span> <span class="mw-headline">Derivation and definition</span></h3>
<p>The work done accelerating a particle during the infinitesimal time interval <i>dt</i> is given by the dot product of <i>force</i> and <i>displacement</i>:</p>
<dl>
<dd><img class="tex" alt="\mathbf{F} \cdot d \mathbf{x} = \mathbf{F} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d (m \mathbf{v})" src="Kinetic_energy_files/5bb75250abd0925c7eb8d60939b9971a.png"></dd>
</dl>
<p>Applying the product rule we see that:</p>
<dl>
<dd><img class="tex" alt="  d(\mathbf{v} \cdot \mathbf{v}) = (d \mathbf{v}) \cdot \mathbf{v} + \mathbf{v} \cdot (d \mathbf{v}) =  2(\mathbf{v} \cdot d\mathbf{v})" src="Kinetic_energy_files/958cb70addede5bb95412c223acd40f6.png"></dd>
</dl>
<p>Therefore (assuming constant mass), the following can be seen:</p>
<dl>
<dd><img class="tex" alt=" \mathbf{v} \cdot d (m \mathbf{v}) = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2  = d \left(\frac{m v^2}{2}\right) " src="Kinetic_energy_files/7a4427b9c8f8d2c4c24792dc9e7f237e.png"></dd>
</dl>
<p>Since this is a total differential (that is, it only depends on the
final state, not how the particle got there), we can integrate it and
call the result kinetic energy:</p>
<dl>
<dd><img class="tex" alt=" E_k = \int \mathbf{F} \cdot d \mathbf{x} = \int \mathbf{v} \cdot d \mathbf{p}= \frac{m v^2}{2} " src="Kinetic_energy_files/3e0c595e5e3bd8f2bcf905acfd6f0da3.png"></dd>
</dl>
<p>This equation states that the kinetic energy (<i>E<sub>k</sub></i>) is equal to the <a href="http://en.wikipedia.org/wiki/Integral" title="Integral">integral</a> of the <a href="http://en.wikipedia.org/wiki/Dot_product" title="Dot product">dot product</a> of the <a href="http://en.wikipedia.org/wiki/Velocity" title="Velocity">velocity</a> (<b>v</b>) of a body and the <a href="http://en.wikipedia.org/wiki/Infinitesimal" title="Infinitesimal">infinitesimal</a> change of the body's <a href="http://en.wikipedia.org/wiki/Momentum" title="Momentum">momentum</a> (<b>p</b>). It is assumed that the body starts with no kinetic energy when it is at rest (motionless).</p>
<p><a name="Kinetic_energy_of_systems" id="Kinetic_energy_of_systems"></a></p>
<h3><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=7" title="Edit section: Kinetic energy of systems">edit</a>]</span> <span class="mw-headline">Kinetic energy of systems</span></h3>
<p>For a single point, or a rigid body that is not rotating, the kinetic energy goes to zero when the body stops.</p>
<p>However, for systems containing multiple independently moving
bodies, which may exert forces between themselves, and may (or may not)
be rotating; this is no longer true.</p>
<p>This energy is called 'internal energy'.</p>
<p>The kinetic energy of the system at any instant in time is simply
the sum of the kinetic energies of the masses- including the kinetic
energy due to the rotations.</p>
<p>An example would be the solar system. In the center of mass frame of
the solar system, the Sun is (almost) stationary, but the planets and
planetoids are in motion about it. Thus even in a stationary <a href="http://en.wikipedia.org/wiki/Center_of_mass_frame" class="mw-redirect" title="Center of mass frame">center of mass frame</a>, there is still kinetic energy present.</p>
<p>However, recalculating the energy from different frames would be
tedious, but there is a trick. The kinetic energy of the system from a
different inertial frame can be calculated simply from the sum of the
kinetic energy in the center of mass frame and adding on the energy
that the total mass of bodies in the center of mass frame would have if
it were moving at the relative speed between the two frames.</p>
<p>This may be simply shown: let V be the relative speed of the frame <i>k</i> from the center of mass frame <i>i</i>&nbsp;:</p>
<dl>
<dd><img class="tex" alt="E_k = \int \frac{v_k^2 dm}{2} = \int \frac{(v_i + V)^2 dm}{2} = \int \frac{(v_i^2 + 2 v_i V + V^2) dm}{2} = \int \frac{v_i^2 dm}{2} + V \int v_i dm + \frac{V^2}{2} \int dm " src="Kinetic_energy_files/5e0f0e623affb579e30d07a8e7d4e9b9.png"></dd>
</dl>
<p>However, let <img class="tex" alt=" \int \frac{v_i^2 dm}{2} = E_i " src="Kinetic_energy_files/b741a4e2b3466e7a8c078fa709ae870e.png"> the kinetic energy in the center of mass frame, <img class="tex" alt=" \int v_i dm " src="Kinetic_energy_files/702d54fa5ecdda24f27a127a10390afa.png"> would be simply the total momentum which is by definition zero in the center of mass frame, and let the total mass: <img class="tex" alt=" \int dm = M " src="Kinetic_energy_files/8d42706bac5ab92d334d4050dff16394.png">. Substituting, we get:<sup id="cite_ref-0" class="reference"><a href="#cite_note-0" title="">[1]</a></sup></p>
<dl>
<dd><img class="tex" alt=" E_k = E_i + \frac{M V^2}{2} " src="Kinetic_energy_files/4582f52032064995bc9bfa1a5cc769b7.png"></dd>
</dl>
<p>The kinetic energy of a system thus depends on the <a href="http://en.wikipedia.org/wiki/Inertial_frame_of_reference" title="Inertial frame of reference">inertial frame of reference</a> and it is lowest with respect to the <a href="http://en.wikipedia.org/wiki/Center_of_mass" title="Center of mass">center of mass</a>
reference frame, i.e., in a frame of reference in which the center of
mass is stationary. In any other frame of reference there is an
additional kinetic energy corresponding to the total mass moving at the
speed of the center of mass.</p>
<p><a name="Rotating_bodies" id="Rotating_bodies"></a></p>
<h3><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=8" title="Edit section: Rotating bodies">edit</a>]</span> <span class="mw-headline">Rotating bodies</span></h3>
<p>If a rigid body is rotating about any line through the center of mass then it has <a href="http://en.wikipedia.org/wiki/Rotational_energy" title="Rotational energy"><i>rotational kinetic energy</i></a> (<span class="texhtml"><i>E</i><sub><i>r</i></sub></span>) which is simply the sum of the kinetic energies of its moving parts, and thus it is equal to:</p>
<dl>
<dd><img class="tex" alt=" E_r = \int \frac{v^2 dm}{2} = \int \frac{(r \omega)^2 dm}{2} = \frac{\omega^2}{2} \int{r^2}dm = \frac{\omega^2}{2} I = \begin{matrix} \frac{1}{2} \end{matrix} I \omega^2 " src="Kinetic_energy_files/77154104d4335916682c1c6f6a1ecac0.png"></dd>
</dl>
<p>where:</p>
<dl>
<dd>ω is the body's <a href="http://en.wikipedia.org/wiki/Angular_velocity" title="Angular velocity">angular velocity</a>.</dd>
<dd>r is the distance of any mass dm from that line</dd>
<dd><i>I</i> is the body's <a href="http://en.wikipedia.org/wiki/Moment_of_inertia" title="Moment of inertia">moment of inertia</a><img class="tex" alt=" = \int{r^2}dm" src="Kinetic_energy_files/695bb453c6ec841b95f9f6fbdbb8f798.png"></dd>
</dl>
<p>(In this equation the moment of inertia must be taken about an axis
through the center of mass and the rotation measured by ω must be
around that axis; more general equations exist for systems where the
object is subject to wobble due to its eccentric shape).</p>
<p><a name="Rotation_in_systems" id="Rotation_in_systems"></a></p>
<h3><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=9" title="Edit section: Rotation in systems">edit</a>]</span> <span class="mw-headline">Rotation in systems</span></h3>
<p>It sometimes is convenient to split the total kinetic energy of a
body into the sum of the body's center-of-mass translational kinetic
energy and the energy of rotation around the center of mass <a href="http://en.wikipedia.org/wiki/Rotational_energy" title="Rotational energy">rotational energy</a>:</p>
<dl>
<dd><img class="tex" alt=" E_k = E_t + E_r \, " src="Kinetic_energy_files/a87463a173b7fc01b280e8b4c6bd1a5b.png"></dd>
</dl>
<p>where:</p>
<dl>
<dd><i>E<sub>k</sub></i> is the total kinetic energy</dd>
<dd><i>E<sub>t</sub></i> is the translational kinetic energy</dd>
<dd><i>E<sub>r</sub></i> is the <i>rotational energy</i> or <i>angular kinetic energy</i> in the rest frame</dd>
</dl>
<p>Thus the kinetic energy of a tennis ball in flight is the kinetic
energy due to its rotation, plus the kinetic energy due to its
translation.</p>
<p><a name="Relativistic_kinetic_energy_of_rigid_bodies" id="Relativistic_kinetic_energy_of_rigid_bodies"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=10" title="Edit section: Relativistic kinetic energy of rigid bodies">edit</a>]</span> <span class="mw-headline">Relativistic kinetic energy of rigid bodies</span></h2>
<p>In <a href="http://en.wikipedia.org/wiki/Special_relativity" title="Special relativity">special relativity</a>, we must change the expression for linear momentum. Integrating by parts, we get:</p>
<dl>
<dd><img class="tex" alt="E_k = \int \mathbf{v} \cdot d \mathbf{p}= \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) = m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} = m \gamma v^2 - \frac{m}{2} \int \gamma d (v^2)" src="Kinetic_energy_files/84d9b73ff32ec72a29aaaf0ab8eadb95.png"></dd>
</dl>
<p>Remembering that <img class="tex" alt="\gamma = (1 - v^2/c^2)^{-1/2}\!" src="Kinetic_energy_files/143039175fcc35f77c6c6e50fd30114e.png">, we get:</p>
<dl>
<dd><img class="tex" alt="E_k = m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d (1 - v^2/c^2) = m \gamma v^2 + m c^2 (1 - v^2/c^2)^{1/2} + C" src="Kinetic_energy_files/05b88b10d6c8bc1a5e1985fdf9eb20d1.png"></dd>
</dl>
<p>And thus:</p>
<dl>
<dd><img class="tex" alt="E_k = m \gamma (v^2 + c^2 (1 - v^2/c^2)) + C = m \gamma (v^2 + c^2 - v^2) + C = m \gamma c^2 + C\!" src="Kinetic_energy_files/4972ca79e8d896a93dccebdb43f6b765.png"></dd>
</dl>
<p>The constant of integration is found by observing that <img class="tex" alt="\gamma = 1\!" src="Kinetic_energy_files/bb02f37d6ee0648401492a5bfb8b669a.png"> when <img class="tex" alt="\mathbf{v }= 0" src="Kinetic_energy_files/636616d0c2410a2240ef1d6d3e94fd4d.png">, so we get the usual formula:</p>
<dl>
<dd><img class="tex" alt="E_k = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - v^2/c^2}} - m c^2" src="Kinetic_energy_files/9a4cbc3faaaa536d8c82fa8921c5e096.png"></dd>
</dl>
<p>If a body's speed is a significant fraction of the <a href="http://en.wikipedia.org/wiki/Speed_of_light" title="Speed of light">speed of light</a>, it is necessary to use <b>relativistic mechanics</b> (the <a href="http://en.wikipedia.org/wiki/Relativity_theory" class="mw-redirect" title="Relativity theory">theory of relativity</a> as expounded by <a href="http://en.wikipedia.org/wiki/Albert_Einstein" title="Albert Einstein">Albert Einstein</a>) to calculate its kinetic energy.</p>
<p>For a relativistic object the momentum p is equal to:</p>
<dl>
<dd><img class="tex" alt=" p = \frac{m v}{\sqrt{1 - (v/c)^2}} " src="Kinetic_energy_files/246efafd67b3f7d269b83b889855d5a9.png">,</dd>
</dl>
<p>where <i>m</i> is the <a href="http://en.wikipedia.org/wiki/Rest_mass" class="mw-redirect" title="Rest mass">rest mass</a>, <i>v</i> is the object's speed, and <i>c</i> is the speed of light in vacuum.</p>
<p>Thus the work expended accelerating an object from rest to a relativistic speed is:</p>
<dl>
<dd><img class="tex" alt="E_k = \frac{m c^2}{\sqrt{1 - (v/c)^2}} - m c^2 " src="Kinetic_energy_files/a234845f2e68dc74e420e32f15d7578d.png">.</dd>
</dl>
<p>The equation shows that the energy of an object approaches infinity as the velocity <i>v</i> approaches the speed of light <i>c</i>, thus it is impossible to accelerate an object across this boundary.</p>
<p>The mathematical by-product of this calculation is the <a href="http://en.wikipedia.org/wiki/Mass-energy_equivalence" class="mw-redirect" title="Mass-energy equivalence">mass-energy equivalence</a> formula—the body at rest must have energy content equal to:</p>
<dl>
<dd><img class="tex" alt="E_\mbox{rest} = m c^2 \!" src="Kinetic_energy_files/a8f5ad763b7c17a4a90ab224a2c05e50.png"></dd>
</dl>
<p>At a low speed (v&lt;&lt;c), the relativistic kinetic energy may be
approximated well by the classical kinetic energy. This is done by <a href="http://en.wikipedia.org/wiki/Binomial_approximation" title="Binomial approximation">binomial approximation</a>. Indeed, taking <a href="http://en.wikipedia.org/wiki/Taylor_expansion" class="mw-redirect" title="Taylor expansion">Taylor expansion</a> for square root and keeping first two terms we get:</p>
<dl>
<dd><img class="tex" alt="E_k \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2\right) - m c^2 = \frac{1}{2} m v^2 " src="Kinetic_energy_files/37ba2311213dd6eec5328d9908474ec7.png">,</dd>
</dl>
<p>So, the total energy E can be partitioned into the energy of the
rest mass plus the traditional Newtonian kinetic energy at low speeds.</p>
<p>When objects move at a speed much slower than light (e.g. in
everyday phenomena on Earth), the first two terms of the series
predominate. The next term in the approximation is small for low
speeds, and can be found by extending the expansion into a Taylor
series by one more term:</p>
<dl>
<dd><img class="tex" alt=" E \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2  + \frac{3}{8} v^4/c^4\right) = m c^2 + \frac{1}{2} m v^2 + \frac{3}{8} m v^4/c^2 " src="Kinetic_energy_files/b8aa721282583f6f75a1fd966e6b7dac.png">.</dd>
</dl>
<p>For example, for a speed of 10 km/s the correction to the Newtonian
kinetic energy is 0.07 J/kg (on a Newtonian kinetic energy of 50 MJ/kg)
and for a speed of 100 km/s it is 710 J/kg (on a Newtonian kinetic
energy of 5 GJ/kg), etc.</p>
<p>For higher speeds, the formula for the relativistic kinetic energy <sup id="cite_ref-1" class="reference"><a href="#cite_note-1" title="">[2]</a></sup> is derived by simply subtracting the rest mass energy from the total energy:</p>
<dl>
<dd><img class="tex" alt=" E_k = m \gamma c^2 - m c^2 = m c^2\left(\frac{1}{\sqrt{1 - (v/c)^2}} - 1\right) " src="Kinetic_energy_files/76cc95f8c7e7d0edde43428308072182.png">.</dd>
</dl>
<p>The relation between kinetic energy and <a href="http://en.wikipedia.org/wiki/Momentum" title="Momentum">momentum</a> is more complicated in this case, and is given by the equation:</p>
<dl>
<dd><img class="tex" alt="E_k = \sqrt{p^2 c^2 + m^2 c^4} - m c^2" src="Kinetic_energy_files/d43af82ca34ad20ae6feea9f6a8d0124.png">.</dd>
</dl>
<p>This can also be expanded as a <a href="http://en.wikipedia.org/wiki/Taylor_series" title="Taylor series">Taylor series</a>, the first term of which is the simple expression from Newtonian mechanics.</p>
<p>What this suggests is that the formulas for energy and momentum are
not special and axiomatic, but rather concepts which emerge from the
equation of mass with energy and the principles of relativity.</p>
<p><a name="Quantum_mechanical_kinetic_energy_of_rigid_bodies" id="Quantum_mechanical_kinetic_energy_of_rigid_bodies"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=11" title="Edit section: Quantum mechanical kinetic energy of rigid bodies">edit</a>]</span> <span class="mw-headline">Quantum mechanical kinetic energy of rigid bodies</span></h2>
<p>In the realm of <a href="http://en.wikipedia.org/wiki/Wave_mechanics" class="mw-redirect" title="Wave mechanics">quantum mechanics</a>, the expectation value of the electron kinetic energy, <img class="tex" alt="\langle\hat{T}\rangle" src="Kinetic_energy_files/9cdb8ad2b590647c9eff7a683cc5d114.png">, for a system of electrons described by the <a href="http://en.wikipedia.org/wiki/Wave_function" title="Wave function">wavefunction</a> <img class="tex" alt="\vert\psi\rangle" src="Kinetic_energy_files/f407865acb630d639a7dd6cf45499c05.png"> is a sum of 1-electron operator expectation values:</p>
<dl>
<dd><img class="tex" alt="\langle\hat{T}\rangle = -\frac{\hbar^2}{2 m_e}\bigg\langle\psi \bigg\vert \sum_{i=1}^N \nabla^2_i \bigg\vert \psi \bigg\rangle" src="Kinetic_energy_files/379828df23a8a61161d3dd01fe007708.png"></dd>
</dl>
<p>where <span class="texhtml"><i>m</i><sub><i>e</i></sub></span> is the mass of the electron and <img class="tex" alt="\nabla^2_i" src="Kinetic_energy_files/2b6cd1cc064daedda6a821242d9ea512.png"> is the <a href="http://en.wikipedia.org/wiki/Laplacian" class="mw-redirect" title="Laplacian">Laplacian</a> operator acting upon the coordinates of the <i>i</i><sup>th</sup>
electron and the summation runs over all electrons. Notice that this is
the quantized version of the non-relativistic expression for kinetic
energy in terms of momentum:</p>
<dl>
<dd><img class="tex" alt="E_k = \frac{p^2}{2m}" src="Kinetic_energy_files/1e8440b3262d1866bf61293634e00f19.png"></dd>
</dl>
<p>The <a href="http://en.wikipedia.org/wiki/Density_functional_theory" title="Density functional theory">density functional</a> formalism of quantum mechanics requires knowledge of the electron density <i>only</i>, i.e., it formally does not require knowledge of the wavefunction. Given an electron density <img class="tex" alt="\rho(\mathbf{r})" src="Kinetic_energy_files/31c9442f7904287e900fc2daa59e6fb6.png">,
the exact N-electron kinetic energy functional is unknown; however, for
the specific case of a 1-electron system, the kinetic energy can be
written as</p>
<dl>
<dd><img class="tex" alt=" T[\rho]  =  \frac{1}{8} \int \frac{ \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r}) }{ \rho(\mathbf{r}) } d^3r " src="Kinetic_energy_files/f527c2275e20ee9262c65527e0d16fe9.png"></dd>
</dl>
<p>where <span class="texhtml"><i>T</i>[ρ]</span> is known as the Von Weizsacker kinetic energy functional.</p>
<p><a name="Some_examples" id="Some_examples"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=12" title="Edit section: Some examples">edit</a>]</span> <span class="mw-headline">Some examples</span></h2>
<p><a href="http://en.wikipedia.org/wiki/Spacecraft" title="Spacecraft">Spacecraft</a> use chemical energy to take off and gain considerable kinetic energy to reach <a href="http://en.wikipedia.org/wiki/Orbital_velocity" class="mw-redirect" title="Orbital velocity">orbital velocity</a>.
This kinetic energy gained during launch will remain constant while in
orbit because there is almost no friction. However it becomes apparent
at re-entry when the kinetic energy is converted to heat.</p>
<p>Kinetic energy can be passed from one object to another. In the game of <a href="http://en.wikipedia.org/wiki/Billiards" class="mw-redirect" title="Billiards">billiards</a>,
the player gives kinetic energy to the cue ball by striking it with the
cue stick. If the cue ball collides with another ball, it will slow
down dramatically and the ball it collided with will accelerate to a
speed as the kinetic energy is passed on to it. <a href="http://en.wikipedia.org/wiki/Collisions" class="mw-redirect" title="Collisions">Collisions</a> in billiards are effectively elastic collisions, where kinetic energy is preserved.</p>
<p><a href="http://en.wikipedia.org/wiki/Flywheel" title="Flywheel">Flywheels</a> are being developed as a method of <a href="http://en.wikipedia.org/wiki/Energy_storage" title="Energy storage">energy storage</a> (see article <a href="http://en.wikipedia.org/wiki/Flywheel_energy_storage" title="Flywheel energy storage">flywheel energy storage</a>).
This illustrates that kinetic energy can also be rotational. Note the
formula in the articles on flywheels for calculating rotational kinetic
energy is different, though analogous.</p>
<p><a name="See_also" id="See_also"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=13" title="Edit section: See also">edit</a>]</span> <span class="mw-headline">See also</span></h2>
<div class="tright portal" style="border: 1px solid rgb(170, 170, 170); margin: 0.5em 0pt 0.5em 0.5em;">
<table style="background: rgb(249, 249, 249) none repeat scroll 0%; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 85%; line-height: 110%;">
<tbody><tr>
<td><a href="http://en.wikipedia.org/wiki/Image:Portal.svg" class="image" title="Portal.svg"><img alt="" src="Kinetic_energy_files/28px-Portal.png" border="0" height="28" width="28"></a></td>
<td style="padding: 0pt 0.2em;"><i><b><a href="http://en.wikipedia.org/wiki/Portal:Energy" title="Portal:Energy">Energy Portal</a></b></i></td>
</tr>
</tbody></table>
</div>
<ul>
<li><a href="http://en.wikipedia.org/wiki/Recoil" title="Recoil">Recoil</a></li>
<li><a href="http://en.wikipedia.org/wiki/Joule" title="Joule">Joule</a></li>
<li><a href="http://en.wikipedia.org/wiki/Parallel_axis_theorem" title="Parallel axis theorem">Parallel axis theorem</a></li>
<li><a href="http://en.wikipedia.org/wiki/Escape_velocity" title="Escape velocity">Escape velocity</a></li>
<li><a href="http://en.wikipedia.org/wiki/Projectile#Typical_projectile_speeds" title="Projectile">Kinetic energy per unit mass of projectiles</a></li>
<li><a href="http://en.wikipedia.org/wiki/Projectile#Kinetic_projectiles" title="Projectile">Kinetic projectile</a></li>
</ul>
<p><a name="Notes" id="Notes"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=14" title="Edit section: Notes">edit</a>]</span> <span class="mw-headline">Notes</span></h2>
<ol class="references">
<li id="cite_note-0"><b><a href="#cite_ref-0" title="">^</a></b> <a href="http://www.phy.duke.edu/%7Ergb/Class/intro_physics_1/intro_physics_1/node64.html" class="external text" title="http://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1/node64.html" rel="nofollow">Physics notes - Kinetic energy in the CM frame</a>. <a href="http://en.wikipedia.org/wiki/Duke" title="Duke">Duke</a>.edu. Accessed 2007-11-24.</li>
<li id="cite_note-1"><b><a href="#cite_ref-1" title="">^</a></b> In Einstein's original <a href="http://www.uni-kiel.de/ub/digiport/ab1800/G4378.html" class="external text" title="http://www.uni-kiel.de/ub/digiport/ab1800/G4378.html" rel="nofollow">Über die spezielle und die allgemeine Relativitätstheorie</a> (Zu Seite 41) and in most translations (e.g. <a href="http://bartleby.com/173/15.html" class="external text" title="http://bartleby.com/173/15.html" rel="nofollow">Relativity - The Special and General Theory</a>) kinetic energy is defined as <img class="tex" alt="m c^2 / \sqrt{1 - v^2/c^2}" src="Kinetic_energy_files/55b60fc5efdf07aca37ae325b9e250b3.png">.</li>
</ol>
<p><a name="References" id="References"></a></p>
<h2><span class="editsection">[<a href="http://en.wikipedia.org/w/index.php?title=Kinetic_energy&amp;action=edit&amp;section=15" title="Edit section: References">edit</a>]</span> <span class="mw-headline">References</span></h2>
<ul>
<li><cite class="book" style="font-style: normal;" id="Reference-Serway-2004">Serway, Raymond A.; Jewett, John W. (2004). <i>Physics for Scientists and Engineers</i>, 6th ed., Brooks/Cole. <a href="http://en.wikipedia.org/wiki/Special:BookSources/0534408427" class="internal">ISBN 0-534-40842-7</a>.</cite><span class="Z3988" title="ctx_ver=Z39.88-2004&amp;rft_val_fmt=info%3Aofi%2Ffmt%3Akev%3Amtx%3Abook&amp;rft.genre=book&amp;rft.btitle=Physics+for+Scientists+and+Engineers&amp;rft.aulast=Serway&amp;rft.aufirst=Raymond+A.&amp;rft.edition=6th+ed.&amp;rft.pub=Brooks%2FCole">&nbsp;</span></li>
<li><cite class="book" style="font-style: normal;" id="Reference-Tipler-2004">Tipler, Paul (2004). <i>Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics</i>, 5th ed., W. H. Freeman. <a href="http://en.wikipedia.org/wiki/Special:BookSources/0716708094" class="internal">ISBN 0-7167-0809-4</a>.</cite><span class="Z3988" title="ctx_ver=Z39.88-2004&amp;rft_val_fmt=info%3Aofi%2Ffmt%3Akev%3Amtx%3Abook&amp;rft.genre=book&amp;rft.btitle=Physics+for+Scientists+and+Engineers%3A+Mechanics%2C+Oscillations+and+Waves%2C+Thermodynamics&amp;rft.aulast=Tipler&amp;rft.aufirst=Paul&amp;rft.edition=5th+ed.&amp;rft.pub=W.+H.+Freeman">&nbsp;</span></li>
<li><cite class="book" style="font-style: normal;" id="Reference-Tipler-2002">Tipler, Paul; Llewellyn, Ralph (2002). <i>Modern Physics</i>, 4th ed., W. H. Freeman. <a href="http://en.wikipedia.org/wiki/Special:BookSources/0716743450" class="internal">ISBN 0-7167-4345-0</a>.</cite><span class="Z3988" title="ctx_ver=Z39.88-2004&amp;rft_val_fmt=info%3Aofi%2Ffmt%3Akev%3Amtx%3Abook&amp;rft.genre=book&amp;rft.btitle=Modern+Physics&amp;rft.aulast=Tipler&amp;rft.aufirst=Paul&amp;rft.edition=4th+ed.&amp;rft.pub=W.+H.+Freeman">&nbsp;</span></li>
<li>School of Mathematics and Statistics, University of St Andrews (2000). <a href="http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Coriolis.html" class="external text" title="http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Coriolis.html" rel="nofollow">Biography of Gaspard-Gustave de Coriolis (1792-1843)</a>. Retrieved on <a href="http://en.wikipedia.org/wiki/2006" title="2006">2006</a>-<a href="http://en.wikipedia.org/wiki/March_3" title="March 3">03-03</a>.</li>
<li><i>Oxford Dictionary</i>, Oxford Dictionary 1998</li>
</ul>


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